Find g' (-1/2), where g(x) is the inverse of f(x) = x3 / (x2 + 1)
g'(x) = 1/f' [g(x)]
The Attempt at a Solution
Everything else I tried got really messy, but I feel like I'm going out on a limb by:
1) y=x3 / (x2+1)
2) y=x3x-2 / 1
4) g '(0.5)= 1/ (0.53 /(0.52+1) = 10
I don't even know if I'm going about it the right way. Any help would be awesome. Thanks.